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3 years ago

What is the difference between a theory and a law?

1 answer:

5 0

A theory is a system of ideas that isn't exactly proven to be true fully. A law is a description of whatever scientific phenomena you're studying. All you need to know is a law describes, and a theory explains.

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How much have jellyfish populated since 2015?

My name is Ann [436]

About a mil sience 2014-2015

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3 years ago

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Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po

gregori [183]

Answer:

E = 0 r <R₁

Explanation:

If we use Gauss's law

Ф = ∫ E. dA = $q_{int}$ / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

E = 0 r <R₁

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3 years ago

A large rocket has a mass of 2.00×10⁶ kg at takeoff, and its engines produce a thrust of 3.50×10⁷ N. Find its initial accelerati

Kazeer [188]

Answer:

17.5 m/s²

1.90476 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Force

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{3.5\times 10^7}{2\times 10^6}\\\Rightarrow a=17.5\ m/s^2$

Initial acceleration of the rocket is 17.5 m/s²

$v=u+at\\\Rightarrow \frac{120}{3.6}=0+17.5t\\\Rightarrow t=\frac{\frac{120}{3.6}}{17.5}=1.90476\ s$

Time taken by the rocket to reach 120 km/h is 1.90476 seconds

Change in the velocity of a rocket is given by the Tsiolkovsky rocket equation

$\Delta v=v_{e}\ln \frac{m_0}{m_f}$

where,

$m_0$ = Initial mass of rocket with fuel

$m_f$ = Final mass of rocket without fuel

$v_e$ = Exhaust gas velocity

Hence, the change in velocity increases as the mass decreases which changes the acceleration

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3 years ago

What are valence electrons?

SashulF [63]

Answer:

The correct answer is D. Electrons in an atom that can bond with other atoms.

Explanation:

For those of you that need it still

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2 years ago

A man releases a stone ar the top edge of a tower during the last second of its travel the stone falls through a distance of (9/

Musya8 [376]

Equations of motion (EoM) use EoM v2=u2+2ax to establish velocities at positions shown in blue in drawing from EoM v=u+at for final 1 second of flight time, we can say v=u+g(1) 2gH−−−−√=2g1625H−−−−−−√+g then, solve for H [in terms of g]

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3 years ago

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